Applied chemistry practical manual session 12 13

(An Autonomous Institute under Rashtrasant Tukadoji
Maharaj Nagpur University, Nagpur)

Practical
Manual
Session: 2012-2013

(An Autonomous Institute under Rashtrasant Tukadoji Maharaj Nagpur University,
Nagpur)

Practical Manual Index Page
Experiment
No.

Aim

Page No.

Group – I
1
2
3
4
5

To find out type of alkalinity and estimate alkalinity present in
the given water sample.
Estimation of temporary, permanent and total hardness present
in supplied hard water sample by complexometric method.
Estimation of copper present in an alloy by iodometry.
Estimation of dissolved Cl2 or residual chlorine content present in
the supplied water sample by Iodometry.
Estimation of Ferrous, Ferric and total iron content present in an
ore by Redox titrimetry.

01
08
19
24
28

Group – II
6

7
8
9
10
11

Determination of relative and kinematic viscosities of a given
lubricating oil at different temperatures using Redwood
Viscometer No. 1 or No. 2.
Determination of flash point of a combustible liquid by Able’s
closed cup flash point apparatus.
Determination of flash point of a combustible liquid by PenskyMartan’s closed cup flash point apparatus.
Determination of flash point of a combustible liquid by
Cleveland’s open cup flash point apparatus.
To determine Chemical Oxygen Demand (COD) of a given water
sample.
To estimate saponification value of a given oil sample.

32

41
46
49
52
57

Applied Chemistry Practical Manual

EXPERIMENT NO. 1

Aim: To find out type of alkalinity and estimate alkalinity present in the given
water sample.

Chemicals: Std. Na2CO3 solution, Unknown water sample A and C, HCl solution (sample B) as an
intermediate solution, phenolphthalein indicator and methyl orange/red indicator.

Principle:
The alkalinity of a water sample is due to carbonates (CO32-), bicarbonates (HCO3-) and
hydroxide (OH-) ions. Thus in a given water sample the possible combination of ions causing alkalinity
are as follows.
(i) OH - alone
(ii) CO3 2 - alone
(iii) HCO3 - alone
(iv) OH- and CO32 – together
(v) CO32-and HCO3- together
The possibility of OH- and HCO3-ions together in the same solution is ruled out as they
react asHCO3 - + OH

-

CO3 2 -- + H2O

In the same way, possibility of co -existence of all the three OH- , CO32-and HCO3- is
ruled out.
The determination of alkalinity involves following reactions.

Shri Ramdeobaba College of Engineering and Management, Nagpur

Page 1


OH- + H+

H2 O

----------------(1)

CO3 2-+ H+

HCO3-

----------------( 2)

HCO3- + H+

H2CO3

H2O + CO2

---------------- (3)

The phenolphthalein is pink in color above pH 10 and it is colorless below pH-8.
While, the methyl orange is yellow above pH 4.4 and it turns pink below pH 3.1
Thus titration of a given water sample in presence of phenolphthalein as an indicator
indicates completion of reaction 1 and 2 whereas the same water sample, if titrated in presence of
Methyl orange as an indicator indicates the completion of reaction 1, 2 and 3.
The water sample, when titrated with an acid solution using phenolphthalein indicator
gives (End point = P) and with methyl orange indicator gives (End point = M). The relation between P
and M points gives the type and extent of alkalinity is established as follows:
Relation between

Extent of alkalinity
Type of alkalinity

OH-

CO32-

HCO3-

P =M

Only OH-

M

-

-

P = 1/2 M

Only CO3 --

-

2P

-

P=0

Only HCO3-

-

-

M

P>1/2 M

OH-&CO3--

(2P-M)

2(M-P)

-

P<1/2 M

CO3-- & HCO3 -

-

2P

(M-2P)

P&M point


Page 2


Procedure:

Part I: - Standardization of HCl solution using standard Na2CO3 solution.

Pipette out 25-ml std. Na2CO3 solution in a 250 ml conical flask. Add 1-2 drops of methyl
orange/red indicator to it. The color of solution turns to yellow. Titrate this reaction mixture with HCl
solution from the burette till the solution color turns to light pink/orange. This is the end point of
titration. Repeat the same procedure to get successive constant end point.

Part II:- Estimation of type and extent of alkalinity present in sample A.

Pipette out 25-ml water sample A in a 250 ml conical flask. Add 1-2 drops of
phenolphthalein as an indicator. The solution becomes pink. Titrate this solution with acid solution
from the burette. At the end point the pink color of the solution changes to colorless. Note this end
point as P point. Now At this point, add 2 drops of methyl orange indicator to the same solution.
Solution becomes yellow. Continue the titration without refilling the burette the solution till the color
of solution turns to orange / pink. Note this end point as M point of the titration. Repeat the same
procedure to get successive constant readings of P and M point.
The relationship between P point and M point denotes the type and extent of alkalinity present in the
given water samples as shown in earlier table.
Part III:- Estimation of type and extent of alkalinity present in sample C.

Pipette out 25-ml water sample C in a 250 ml conical flask and follow the same procedure as given in
part II.


Page 3


Observation table:
Part I: - Standardization of HCl solution using standard Na2CO3 solution.
Standard Na2CO3 solution Vs HCl solution
Control reading:

to

ml

Volume of HCl
S.No

Vol. of Na2CO3

Constant Reading

solution

solution (V1 ml)

(V2 ml).
IBR

FBR

Part II:- Estimation of type and extent of alkalinity present in sample A.
Given water sample A Vs HCl solution
Control reading:

P point:

M Point: :

S.No

Vol. given water
sample A (V2 ml)

Volume of HCl solution (ml)

to

to

ml
ml

End
Point

IBR

P point

M point

(ml)
P point=
M point=


Page 4


Part II:- Estimation of type and extent of alkalinity present in sample C.
Given water sample C Vs HCl solution
Control reading:

P point:

M Point: :

S.No

Vol. given water
sample C (V3 ml)

Volume of HCl solution (ml)

to

to

ml
ml

End
Point

IBR

P point

M point

(ml)

P point=
M point=

Calculations:Part I: - Standardization of HCl solution using standard Na2CO3 solution.

Given: Normality of Na2CO3 solution = N1

Standard Na2CO3 solution

HCl solution

N1 X V1

=

N2 X V2
N1 X V1

N2

=

-----------------V2

The given Normality of HCl solution is = N2


Page 5


Part II : Estimation of type and extent of alkalinity present in sample A
For sample A, P =

ml and M =

ml

If P > 1 / 2 M
then,
The type of alkalinity present in given water sample A is OH- and CO32Now, 1)Volume of acid corresponding to OH - = 2P- M
2) Volume of acid corresponding to CO32- = 2 (M- P)

a) Normality of water sample due to OHHCl

Water sample A

N2V2

=

N3V3

N2 x 2P- M
N3 =

------------------25

Wt/ lt for OH- = Normality x Eq. Wt.of CaCO3
=

N3

=

x

50
g/l

=
=

x 1000 mg/l
ppm


Page 6


b) Normality of water sample due to CO3 2-.
HCl

Water sample A

N2V2

=

N4V4
2 (M- P)

N2 X
N4

= ------ -------------------25

Wt./ lt for CO3 2-.

=

Normality

=

x Eq.. Wt. of CaCO3
x

N4

=

50 g/l

x 1000 mg/l

=

ppm

Same calculations has to be carried out for Water sample C

Result: The given water sample A contains:OH – Alkalinity

= ------------ ppm CaCO3 equivalent.

CO3 2– Alkalinity

= ------------- ppm CaCO3 equivalent.

Total Alkalinity


The given water sample C contains:HCO3 – Alkalinity

= ------------ ppm CaCO3 equivalent.

CO3 2– Alkalinity


Total Alkalinity



Page 7


EXPERIMENT NO. 2
Aim : Estimation of temporary, permanent and total hardness present in supplied
hard water sample by complexometric method.

Chemicals Required: Standard hard water (sample A), given water sample (Sample C), ethylene
diamine tetra-acetic acid solution as an Intermediate solution (Sample B), Buffer solution (NH4Cl +
NH4OH having pH = 10), Eriochrome Black-T indicator solution.

Principle: Hard water contains the dissolved salts of calcium, magnesium and iron ions which are
called hardening ions. In low concentrations these ions are not considered harmful for domestic use,
but at higher concentrations of these ions interfere with the cleansing action of soaps and accelerate the
corrosion of steel pipes, especially those carrying hot water.

Hardening ions, such as Ca2+ and Mg2+, form insoluble compounds with soaps. Soaps, which
are sodium salts of fatty acids such as sodium stearate, C17H35CO2Na, are very effective cleansing
agents so long as they remain soluble; the presence of the hardening ions however causes the formation
of a gray, insoluble soap scum such as (C17H35CO2)2Ca.

Groundwater becomes hard as it flows through underground limestone (CaCO3) deposits;
generally, the water from deep wells has a higher hardness than that from shallow wells because of a
longer contact time with the limestone. Surface water similarly accumulates hardening ions as a result
of flowing over limestone deposits. In either case the CO2 dissolved in rainwater solubilizes limestone
deposits.

Page 8


Hard water is also responsible for the appearance and undesirable formation of "boiler scale".
The boiler scale is a poor conductor of heat and thus reduces the efficiency of transferring heat.
Experimental background:
Complexometry involves the interaction of an organic compound with a complexable metal ion and
results in the formation of compounds known as Werner’s complexes. The organic compound is an anion
or a Lewi’s base and is known as Ligand. Several factors, such as pH, basicity of ligand, type of ring
formed etc. govern the successful formation of a complex. Thus:

XL
Ligand

+ M n+

--------- (L) x M

Metal ion

Complex

In this experiment a titration technique is used to measure the combined Ca2+ and Mg2+ concentrations
in a water sample. The titrant or the intermediate solution is the disodium salt of ethylene diamine
tetraacetic acid (abbreviated Na2H2Y or EDTA). Its structure is as below:

NaOOC-H2C

CH2-COOH

/
N-CH2-CH2-N

/
HOOC-H2C

CH2COONa

Disodium salt of EDTA
In aqueous solution Na2H2Y dissociates into Na+ and H2Y-2 ions. The H2Y-2 ion reacts with the
hardening ions, Ca2+ and Mg2+, to form very stable complex ions, especially in a solution buffered at a

Page 9


pH of about 10. As Ca2+ and Mg2+ is present in the hard water and EDTA solution are adding from the
burette, it complexes with the "free" Ca2+ and Mg2+ of the water sample to form the respective complex
ions:

From the balanced equations, once the molar concentration of the Na2H2Y solution is known, the
moles of hardening ions in a water sample can be calculated from a 1:1 stoichiometric ratio. Thus it is
evident that 1 mole of the complex forming H2 Y2 - reacts in with one mole of the metal ion and, two
moles of hydrogen ions are generated. It is apparent from the equation above that the dissociation of
the complex will be governed by the pH of the solution. Lowering of the pH will decrease the stability
of the metal EDTA complex. Thus, pH should be maintained constant by the addition of basic buffer in
the reaction mixture.
The structure of EDTA-metal complex is as follows:

CH2-COO-----M-------OOC-CH2

M = Ca2+, Mg2+

/
N---------CH2-CH2--------N

/
CH2-COO-

-COOCH2

An indicator is used to detect the endpoint in the titration, Eriochrome Black T (EBT). Eriochrome
black-T is sodium 1 – (1-hydroxy, 2- napthyl azo ) 6-nitro – 2 napthol-4, sulphonate.


Page 10


HO

OH

N

NaO 3S

N

O2N

Hence EBT for the simplicity is represented asNaH2D which gives H2D – ion which exhibits different
colours at different pH values.

H2D - ----------- H D 2- -------------
pH 6.3

pH 10

( Red)

D3pH 11.5

(Blue)

(Yellowish Orange)

In the pH range 8-10, the blue form of the indicator HD 2- gives a wine red complex
with Metal ion.
For ex.:
Mg 2+ + HD2- ---------- Mg D- + H+
Blue

Wine red

It forms complex ions with Ca2+ and Mg2+ ions, but binds more strongly to Mg2+ ions.
In this estimation, thus four types of complexes are possible
viz :


Page 11


a) EBT – Mg2+ complex

(Wine red)

b) EBT - Ca2+ complex

(Wine red)

c) EDTA- Mg2+ complex

(colourless)

d) EDTA- Ca2+ complex

(colourless)

Their order of stability and consequently, their preference of formation are:

EDTA- Ca2+ > EDTA- Mg2+ > EBT – Mg2+ > EBT – Ca 2+

Because only a small amount of EBT is added, only a small quantity of Mg2+ is complexed; no
Ca2+ ion complexes to EBT — therefore, most of the hardening ions remain "free" in solution. The
EBT indicator is sky-blue in solution but forms a wine-red complex ion, [Mg-EBT] 2+.

pH
EBT + Mg2+ -------- EBT-Mg2+ Complex
8-10

(Wine red colour obtained)

Before any EDTA solution is added for the analysis, the reaction mixture is wine-red because
of the [Mg-EBT]2+ complex ion. Once the EDTA complexes all of the "free" Ca2+ and Mg2+ from the
water sample, it then removes the trace amount of Mg2+ from the wine-red [Mg-EBT] 2+ complex ion.
The solution changes from wine-red back to the sky-blue color of the EBT indicator, and the endpoint
is reached — all of the hardening ions have been complexed with EDTA. The reaction is given as
follows:


Page 12


pH
EDTA + Ca2+ -------- EDTA-Ca2+ complex (Colourless)
8-10
pH
EDTA + Mg2+ -------- EDTA- Mg2+ complex (Colourless)
8-10
pH
EDTA + EBT-Mg 2+ complex
(Wine red)

---------- EDTA-Mg 2+ complex

+

8-10

EBT
(blue)

PROCEDURE:
Part A: Standardization of EDTA solution using standard hard water sample
Pipette out 25 ml. of standard hard water solution in a clean conical flask. To it add 5 ml
buffer solution (NH4Cl + NH4OH) and 3-4 drops of internal indicator EBT. The whole solution
becomes wine-red in colour. Titrate this solution against EDTA solution from the burette. A change of
colour of the solution from wine-red to blue indicates end point. Repeat the same procedure to get
successive constant reading.
Part B: Estimation of total hardness present in given water sample C
Pipette out 25 ml. of the hard water sample C into a 250 ml. conical flask. To it add 5
ml buffer solution (NH4Cl + NH4OH) and 3-4 drops of internal indicator EBT. The whole solution
becomes wine-red in colour. Titrate this solution against EDTA solution from the burette. A change of
colour of the solution from wine-red to blue indicates end point. Repeat the same procedure to get
successive constant reading.


Page 13


Part C: Estimation of Parmanant hardness present in given water sample C
Pipette out 100 ml of given hard water sample C in 250 ml borosil conical flask. Boil this
solution gently to reduce to 1/3rd of its original volume. Filter this warmed solution into 250 ml
volumetric flask. After complete transfer of boiled sample, remove the filter paper. Wash the funnel
properly with distilled water and collect the washings in the volumetric flask. Then make up the
volume up to the mark of volumetric flask using distilled water. Shake the solution and then transfer it
to 250 ml clean beaker.
Now pipette out 25 ml of diluted hard water sample C and add 5 ml buffer solution
(NH4Cl + NH4OH) and 3-4 drops of internal indicator EBT to it. Titrate this reaction mixture with
EDTA solution from the burrete until the wine red colour changes into blue. Repeat the same
procedure to get successive constant reading.

Observation Table:
Part 1: Standardization of EDTA solution using standard hard water sample
Standard Hard Water Sample Vs EDTA solution
Control reading:

to

ml

Volume of EDTA
S.No

Vol. of std. Hard

Constant Reading

solution

water (V1 ml)

(V2 ml).
IBR

FBR


Page 14


Part 2: Estimation of Total Hardness present in given water sample C.
Water Sample C Vs EDTA solution
Control reading:
Vol. of given
S.No

Hard water
sample C (V3 ml)

to

ml

Volume of EDTA
Volume of EDTA

solution

solution
IBR

IBR

Diluted Water Sample C Vs EDTA solution
Control reading:
Vol. of Diluted
S.No

Hard water
sample C (V3 ml)

to

ml

Volume of EDTA
Volume of EDTA

solution

solution
IBR

IBR


Page 15


Calculations:
For part 1: Standardization of EDTA solution using standard hard water sample
Let W g of hardness causing salt is dissolved in y ml of water.
W X 1000
Normality of std. Hard water = -------- -------------------------------------- = M1
Eq. wt. of salt

X y

Std Hard water

EDTA

N1 X V1

=

N2 X V2

N1 X V1
N 2 = -----------------V2
= (Normality of EDTA)
Part 2 : Determination of Total Hardness present in given water sample.
EDTA

Water Sample

N2 X V2’

=

N3 X V3
N2 X V2’

N3

=

-----------------V3


Page 16


= (Normality of hard water)
Therefore the amount of Total Hardness present in given water sample can be calculated as:
Wt/Lit

= N3 x Equivalent Wt. of CaCO3 X 1000
=

A

mg/ lit CaCO3 equivalent hardness.

Diluted Water Sample C

EDTA solution

N2 X V2”

=

N3 X V3
N2 X V2”

N3

=

-----------------V3

= (Normality of diluted hard water)
Therefore the amount of Permanent Hardness present in diluted water sample can be calculated as:

Wt/Lit

= N3 x Equivalent Wt. of CaCO3 X 1000
= B

mg/ lit CaCO3 equivalent hardness.

This much amount of hardness is present in diluted water sample C.
But, 250 ml diluted water sample C is prepared from 100 ml water sample C i.e. 2.5 times diluted.
Hence, permanent hardness in water sample C = B X 2.5 mg/ lit CaCO3 equivalent hardness.
Therefore, Temperory hardness present in given water sample C
= (A-B) mg/ lit CaCO3 equivalent hardness


Page 17


Result: The amount of hardness present in given sample C is as follows:
Permanent Hardness =

mg/ lit CaCO3 equivalent hardness

Temporary Hardness =


Total Hardness


=

EXPERIMENT NO. 3


Page 18


AIM: Estimation of copper present in an alloy by iodometry.
THEORY: The Iodometric method of volumetric analysis is based on oxidation -reduction process
(Redox) and involves interconversion of elemental Iodine into Iodide ions and vice-versa asIodometry
2 I - -------------- ---------- I2 + 2eIodimetry
There are two types of Iodometric titrations
a) Indirect Iodometric titrations (Iodometry) : It involves titration of liberated iodine from
potassium iodide (KI) in a chemical reaction.
b) Direct Iodometric titrations (Iodimetry) : It involves titration with Standard solution of Iodine.

PRINCIPLE:In this Experiment, an oxidizing agent such as K2Cr2O7 or CuSO4 is treated in acid solution (dil.
H2SO4) with an excess of iodide ion produced by KI. The oxidizing agent oxidizes the iodide ions into Iodine,
which is then titrated with the standardised solution of reducing agent such as sodium thiosulphate solution.
(Hypo)

Chemical Reactions:
PART A: Standardization of Sodium thiosulphate solution.
1)

K2Cr2O7 + 6KI + 7H2SO4

3I2 + Cr2(SO4)3 + 4K2SO4 + 7H2O

In ionic form, the same reaction can be written as


Page 19


Cr2O72- + 6I- + 14H+

3I2 + 2Cr3+ + 7H2O

Thus the amount of Iodine liberated depends only upon the quantity of K2Cr2O7 act as limiting
reagent.
2)

2 Na 2S2O3 +

I2

2Na I +

Sodium thiosulphate

Na 2 S4 O6

Sodium tetrathionate

2S2O32- + I2

2I- + S4O62-

PART B: Estimation of copper ions:
2CuSO4 +

4KI

Cu2I2 + I2 +2 K2SO4

2Cu2+ + 4 I-

2CuI2

Cu2I2 + I2

The amount of iodine thus generated depends upon the concentration of Cu2+ ions which can be
estimated by titrating it against sodium thiosulphate solution using starch as an indicator.
Since the solubility of Iodine in water is low, a considerable excess of KI must be used in
iodometric determination of oxidizing agent. The iodine liberated by the reaction then dissolves by
forming the complex K (I3)
KI + I2

K (I3)

OR

I-+ I2

(I3)-

Formation of this compound (I3)- does not interfere with in the titration of iodine with Sodium
thiosulphate because, as the free iodine reacts with thiosulphate, the equilibrium between I2 and
(I3)- is disturbed and more I2 is produced by the backward reaction.
After addition of KI and /or acid the flask is kept in dark place for about 5 minutes
as rate of the reaction between oxidizing agent and I- is too low.. Secondly, light accelerates side
reaction of photo-oxidation as,


Page 20


4I-+ 4H+ + O2

2I2 + 2H2O

Here, I- are oxidized to I2 by atmospheric oxygen and analytical results may go
wrong. It is for this purpose ,before adding KI to the acidic solution of the oxidizing agent, a pinch
of Na2CO3 is added so as to generate a CO2 atmosphere which would prevent access of the O2 to
the solution.
Indicator:
Most commonly used indicator in iodometric titration is STARCH. It forms deep
blue complex with iodine (but not with iodide)
Starch is a polymer, consisting of two components Amylose and
Amylopectine. The active fraction amylose is a polymer of the sugar α –D- glucose. It has a shape
of coiled helix into which I2 combined with I- can fit, thus reacting slowly with thiosulphate. Due to
this reason the starch solution must be added at the very end of the titration (when iodine conc. is
very less).
Starch solution is not very stable solution because various microorganisms are
degrading it and one of the degradation products, glucose, is a reducing agent. Thus an old starch
solution containing some glucose can cause an appreciable titration error. Hence always fresh
starch solution is used.

PROCEDURE: Part A: -Standardization of sodium thiosulphate solution.
Pipette out 25ml. std. K2Cr2O7 solution in a conical flask. Add 10 ml.dil. sulphuric acid, followed by
a pinch of sodium bicarbonate /sodium carbonate and finally 10 ml KI (10%). Keep the flask covered
for 5-6 minutes in dark. Appearance of dark brown color indicates the formation of I2.


Page 21


Now, titrate it against Na2S2O3 from burette until dark brown colour becomes faint brown. At this
stage, add 2 ml. starch solution, colour of the solution becomes intense blue. Continue titration drop
wise till blue color disappears. Note this point as an end point.
Part B: -Estimation of Cu2+ ions.
Pipette out 25 ml. given CuSO4 solution in a conical flask. Add few drops of dil. Na2CO3
solution until a faint permanent precipitate remains. Add acetic acid solution dropwise till the solution
becomes clear. Now add 10 ml. KI (10%). Appearance of dark brown colour indicates formation of I2.
Titrate the liberated iodine with Na2S2O3 solution from burette till brown colour of iodine fades, then
add 2ml. of starch solution, and continue the addition of the Na2S2O3 solution until the blue color
commences to fade. Then add about 5-6ml of ammonium thiocyanate solution into the flask due to
which blue color become more intense. Continue the titration till blue color disappears (there is
formation of flesh coloured ppt in the flask). Note this point as an end point.

Calculations:-

Given Normality of std. K2Cr2O7 is N1
Part A: -Standardization of Sodium thiosulphate solution.(Hypo.)
Std. K2Cr2O7

v/s

Hypo

N1V1

=

N2V2

N1x 25

=

N2 X V2
N1x 25

=

N2

----------------V2

N2

=

X

(Normality of hypo solution)


Page 22


Part B: -Estimation of Cu2+ions.
Hypo

v/s

N2V’2

given CuSO4
=

N3V3
N2V’2

N3 =

-----------25

N3 =

Y

(Normality of CuSO4 solution.)

= Normality x Eq. Wt. Of Cu2 + ions of Cu 2 +

Wt./lt.of
=

N3
=

X

63.54

Z g/l

Result: Hence, The given solution of copper salt was found to contain Z g/l of Cu2+ ions.
Precautions:
1) To suppress the volatility of Iodine, excess of KI is added.
2) Flask is to be kept in dark to avoid photo oxidation of Iodide ions to Iodine.
3) Freshly prepared starch solution only can be used as an indicator.


Page 23


EXPERIMENT N0. 4
Aim: Estimation of dissolved Cl2 or residual chlorine content present in the
supplied water sample by Iodometry.
Theory:
Living organisms such as algae, fungi and bacteria are more abundant in surface drainage
waters while in deep well waters, the bacterial count is often low .The growth of these organisms in
water may have serious consequences .The slime surrounding these organisms causes them to adhere
to the metal surfaces and the film of slime thus formed serves as a base to which suspended matter
present in the water can adhere. This leads to a serious reduction in the heat transfer rates and tube
blockages. The growth of marine organisms such as mussels may lead to serious reduction in the
carrying capacity of pipelines and culverts. Thus measures to prevent the development of the living
organisms are often necessary. Organism growth generally take place most readily in water at
temperature range between 10 - 35 0C.
Chlorine is widely used for disinfecting water to remove bacteria, fungus and other
pathogenic microorganisms and for deodorization, as it is a powerful oxidizing agent and is cheaply
available.
Chlorination can be carried out by using bleaching powder or chlorine gas or chlorine
dissolved in water in the form of concentrated solution or with chloramines. The sterilizing action of is
supposed to be due to its reaction with water producing hypochlorus acid and nascent oxygen both of
which have powerful germicidal properties.
Reactions:
CaOCl2
Cl2 +

+ H2 O
H2O
HOCl

Hypochlorus acid

Ca (OH) 2 + Cl2
HOCl + HCl
HCl

+ [O]
nascent oxygen


Page 24


Principle:
The principle involved is, when a measured quantity of chlorinated water is treated with
excess of potassium iodide, the free chlorine present in the water oxidizes the corresponding amount of
potassium iodide to Iodine. The liberated iodine is estimated by titrating against sodium thiosulphate
solution.
Cl2

+

2KI

2KCl

I2 + 2Na2S2O3

+

I2

Na2S4O6 + 2NaI

Sodium thiosulphate

sodium tetra thionate

Reagents required
1. Std K2Cr2O7 Solution
2. KI solution (10%)
3. Conc. HCl
4. NaHCO3
5. N/20 Hypo solution

6. Starch solution

Procedure:
Part A: - Standardization of Hypo solution
Pipette out

25ml. std. K2Cr2O7 solution in a conical flask. Add 10 ml.dil. sulphuric acid,

followed by a pinch of sodium bicarbonate /sodium carbonate and finally 10 ml KI (10%). Keep the
flask covered for 5-6 minutes in dark. Appearance of dark brown color indicates the formation of I2.
Now, titrate it against Na2S2O3 from burette until dark brown colour becomes faint brown. At
this stage, add 2 ml. starch solution, colour of the solution becomes intense blue. Continue titration
drop wise till blue color disappears. Note this point as an end point.
Part B: - Determination of chlorine in the given water sample.
Take 25-ml.given water sample in a conical flask. Now add 10 ml. KI (10%). Appearance of
dark brown color indicates formation of I2. Titrate the liberated iodine with Na2S2O3 solution from


Page 25


burette till brown color of iodine fades, then add 2-ml. of starch solution, and continue the addition of
the Thiosulphate solution until the blue color disappears completely. Note this point as an end point.

Calculation:Given Normality of std. K2Cr2O7 is N1
Part A: -Standardization of Sodium thiosulphate solution (Hypo.)

Std. K2Cr2O7

v/s

Hypo

N1V1

=

N2V2

N1x 25

=

N2 X V2

N1 x 25
N2

= ----------------V2
=

X (Normality of Hypo solution)

Part B: - Determination of chlorine in the given water sample.

Hypo
N2V’2

v/s

given water sample
=

N3V3
N2V’2

N3 =

-----------25

N3 =

Y (Normality of given water sample.)


Page 26


Wt./lt. of Chlorine =

Normality x Eq. Wt. of Cl 2
=

N3

=

Z g/l

X

35.5

Result:- Hence, The amount of free residual chlorine found to be present in the given
chlorinated water sample is -----------ppm


Page 27


EXPERIMENT N0. 5
Aim: - Estimation of Ferrous, Ferric and total iron content present in an ore by
Redox titrimetry.

Principle:
The principle involved in this titration is that only Ferrous ions gets oxidized to ferric ions and
the ferric ions in the solution remains as it is.
The amount of ferrous ions present in the given solution containing ferrous and ferric
salt is determined by titrating the solution, and then the total iron (present now in the ferrous state
only) is titrated with potassium dichromate.
1) Cr2O72- + 6Fe2+ +14H+

2Cr3+ + 6Fe3+ + 7H2O

2) 2Fe3+

2Fe 2+ + Sn 4+

+ Sn 2+

In acid solution reduction of potassium dichromate may be represented as:
Cr2O7 2- + 14H + + 6e-

2Cr3 +

+ 7H2O

.
Ferric salts are not oxidized either by dichromate or permanganate. They can be
determined in solution only after reduction of Fe3+ to Fe2+ ions. This reduction can be carried out by
reducing agents such as SnCl2. Ferric salt e.g. FeCl3 is reduced by the action of SnCl2 in presence of
hydrochloric acid.

Fe2 (SO4) 3 + 6HCl
FeCl3 +

SnCl2

6FeCl3 + 3H2SO4
2FeCl2 + SnCl4

The excess SnCl2 must be removed completely, as K2Cr2O7 oxidizes it. This is done by
adding mercuric chloride, which reacts with SnCl2 as follows:


Page 28


SnCl2 + 2HgCl2
(Excess added)

SnCl4 +

Hg2Cl2

(Silky white ppt.)

Mecurous chloride (calomel Hg2Cl2) is deposited in the form of silky white
precipitate. This is not filtered off and the solution is directly titrated with K2Cr2O7, Following reaction
takes place: -

6FeCl2 + K2Cr2O7 + 14HCl

6FeCl3 + 2KCl + 2CrCl3 + 7H2O

The following points must be taken into consideration in this analysis. The
precipitated Hg2Cl2 can also oxidize by dichromate. However, if only a small excess SnCl2 is used in
the previous stage, the amount of HgCl2 precipitated is small, it reacts with the dichromate slowly and
no appreciable error is introduced into the result. On the other hand, if a large excess of Hg2Cl2 is
precipitated and, in particular, if it has a gray or dark colour as a result of further reduction to metallic
mercury by the reaction

Hg2Cl2 + SnCl2

2Hg + SnCl4

Then it is oxidized so rapidly in the titration that the result is quite erroneous.

It follows, therefore, that a basic condition for accuracy in this method is the use of a
very slight excess of SnCl2. Stannous chloride is added drop by drop until the reddish yellow/ yellow
colour of FeCl3 has completely disappeared, after which one or two more drops are added.

PROCEDURE:

Part-I : Estimation of Ferrous ions present in given mixture:
Pipette out 25 ml.of the given Ferrous & Ferric salt solution mixture in a conical flask, add
about 10 ml. of sulphuric acid and 5-6 drops of N-Phenyl anthranilic acid as an indicator. Titrate


Page 29


slowly with constant shaking of the flask against K2Cr2O7 solution till the brick red colour, which
marks the end point. This titre value corresponds to the ferrous ions present in the solution.

Part-II : Determination of total iron content present in given mixture.

Pipette out 25 ml.of the given Ferrous & Ferric salt solution mixture in a conical flask and
add 10 ml. of conc. HCl, The solution turns reddish yellow due to the formation of ferric chloride. Heat
the flask and to the hot solution (70-900C) add concentrated stannous chloride solution (SnCl2) drop
wise with constant stirring till the solution becomes colourless. Add one drop extra to ensure complete
reduction.
Cool the solution rapidly under the tap water.Add 5 ml of Mercuric chloride at once. A
silky white precipitate of mercurous chloride is formed. [However, if a heavy precipitate forms, or a
gray or black precipitate is obtained, it means that too much of SnCl2 has been used. If no precipitate
was formed, it shows that SnCl2 added was insufficient. In either of these above cases, the expt. has to
be rejected and fresh start should be made. ]

To this white ppt containing solution, add 4-5 drops of N-Phenylantranilic acid as an
indicator.Titrate it against std. K2Cr2O7 solution,till the solution turns brick red. Mark this point as end
point.

CALCULATIONS:
Part-I : Estimation of Ferrous ions present in given mixture:

Std. K2Cr2O7 vs
N1V1

=

Fe2+ / Fe3+ mixture
N2V2
N1V1

N2

=

---------------V2

N2

= ------------(Normality

of Solution w.r.to Fe2+)


Page 30


Wt./ lit.of Fe2+ =
=

N2

x

X

55.85

g/l

Part-II : Determination of total iron content present in given mixture.
Fe2+ / Fe3+ mixture

Std. K2Cr2O7 vs

N1V1

=

N3V3
N1V1

N3

=

--------------V3

N3
Wt./ lit.of

=

-----

= N3

g/l (Normality w.r.to Total iron content)

x

55.85

= Y g/l
So, The amount of Fe3+ ions present in the mixture = (Y – X) g/l

.

Result: The amount of Fe2+, Fe3+ and total iron content present in given mixture was found to be
respectively.


Page 31


EXPERIMENT NO. 6
Aim: Determination of relative and kinematic viscosities of a given lubricating oil
at different temperatures using Redwood Viscometer No. 1 or No. 2.
Apparatus: Redwood viscometer (1 / 2), thermometer, etc.

THEORY AND GENERAL DISCUSSION:

Viscosity is one of the most important properties of a lubricating oil. Viscosity is a measure
of the internal resistance to the motion of a fluid and is mainly due to the forces of cohesion between
the fluid molecules. The formation of a fluid film of a lubricant between the moving friction surface
and the generation of the frictional heat under the particular condition of load, bearing speed and
lubricant supply is mainly depends upon the viscosity of the lubricant and to some extent on its
oiliness. If the viscosity of the oil is too low, the fluid lubricant film cannot be maintained between the
moving surfaces as a result of which excessive wear may take place. On the other hand, if the viscosity
of the oil is too high, excessive friction due to the shearing of oil itself would result. Hence in
hydrodynamic lubrication, the lubricant should posses the proper viscosity. So, it is of vital importance
to have knowledge of the viscosity of the lubricating oil. Viscosity of the fluid may be measured in
several ways, one of which is determining time required for a definite amount of the liquid to flow
through a capillary. Such method includes the use of Saybolt, Engler and Redwood Viscometers.

Viscosity may be expressed as dynamic (or absolute) viscosity, kinematic viscosity or the
viscosity called after the name of the apparatus used for its determination.

DYNAMIC OR ABSOLUTE VISCOSITY:
Dynamic or absolute viscosity (often denoted by ‘n’) is the tangential force per unit area
required to maintain unit velocity gradient between two parallel planes, in the fluid unit distance apart.
It can be also be defined as the ratio of shearing stress to the rate of shearing strain.


Page 32


If F is the force required to keep moving an particle of surface area A in contact with the
fluid, separated from stationary surface by a thickness D, and moving with velocity V, then

Shearing stress = Force acting on the surface area A
i.e. F/A: ---------------------------(1)

Rate of shearing strain = V/D ----------------------------(2)

Absolute Viscosity n

F/A

= Shearing stress/ rate of shearing strain.

FxD

------------ = ------------V/D

-----------(3)

VxA

The numerical value of ‘n’ depends upon the unit used in equation (3). In metric system
the unit is ‘poise’ (F = 1 dyne, A = 1 cm. And V = km/sec.). A smaller unit, the centipoises is more
often used. Poise is equal in one dyne/second/cm2 (Dimensions of absolute viscosity are ML

–1 -1

T ).

Absolute is also referred to as ‘coefficient of viscosity’.
KINEMATIC VISCOSITY: The ratio of absolute viscosity to density for any fluid is known as absolute
kinematics viscosity. It is donated by µ and in C.G.S. system, its units are stokes and centistokes
(1/100 th of stoke) respectively.
µ = η /p
Where,
µ = Absolute kinematic viscosity
ŋ = absolute dynamic viscosity


Page 33


ρ = density of fluid.
Dimension of µ are L2 T1
Since the rate, at which a fluid will flow through an apparatus increases as the internal
friction of the fluid decreases, the rate of flow through an orifice or short tube may be used as a means
for measuring viscosity. This is the principle involved in the Redwood viscometer, which is an English
standard whereas Saybolt’s Viscometer is used in U.S.A. and Engler’s Viscometer in Europe. In these
commercial Viscometers a fixed volume of the liquid is allowed to flow (in case of Redwood it is 50
ml. at 270C Engler-60 ml. and Saybolt –200 ml.) through a capillary tube of specified dimension under
given set of conditions and the time of flow is measured at a particular temperature. The result is
usually expressed in terms of the time in seconds taken by oil to flow through the standard orifice of
the particular standard apparatus used.
e.g.: Viscosity of oil is 250, Redwood ( no. ½) seconds at 270C. This Viscosity so
determined is sometimes called as Relative Viscosity.
Absolute and kinematic Viscosities can also be determined from the relative Viscosity
(i.e. Redwood values) from the equations:
µ = Ct

----------------(5) (for Fluids)

Whose Kinematic Viscosity is more than 10 centistokes
And m = Ct – β / t ----------------(6) (for fluids Kinematics Viscosities less than or
Equal to 10 centistokes)
Where,
µ = kinematic viscosity in centistokes
t

= time of flow in seconds

C = viscometer constant


Page 34


ρ

= coefficient of kinetic energy which may be determined experimentally or
eliminated by choosing long flow-times

Test Viscometer may be calibrated constant C determined by solutions of known Viscosity.
The primary standard used is freshly distilled water, whose kinematics viscosity is 1.0008
centistokes. Other standards usually employed are40% sucrose solution
µ = 4.390 cs at 25 0C

p = 1.1739

60% sucrose solution.
µ = 33.66 cs. At 250C

p = 1.28335

For Redwood viscometer No.1, the values for the constants are as below:
Time of flow, t
40 to 85 seconds
85 to 2000 seconds

β
190
65

C
0.264
0.247

For Redwood viscometer No.2, the values for the constants are β =1120, C= 2.720
Redwood viscometer No. 2 is used for very viscous liquids and
gives 1/10 th the value of Redwood No.1 viscometer. Replace the ball value in position to seal the cap
to prevent overflow of the oil.


Page 35


Diagram:

PROCEDURE:
1) Level the instrument by leveling screws ensuring that it is horizontal
With the help of spirit level.
2) Put the valve rod at its position (i.e. in the concavity at the bottom of
The cup) to close the passage of orifice.
3) Keep clean dry Kohlrausch’s flask centrally below the jet.
4) The test sample is poured into the oil cup and adjust the level of oil,
with that of the pointer. Later indicates the level to which oil is to be


Page 36


filled.
5) Fill up the water bath with water and adjust the level of water with that
of the pointer in the oil cup.
6) Cover the oil cup with the lid and insert thermocouple / thermometer
of desired range into the oil cup from the thermometer bracket. Care
should be taken that it dose not touch the bottom of the cup.
7) Also insert the thermometer into the water bath from the shield.
8) Let the oil in the cup as well as water of the water bath attain the
room temperature . Once the temperature of oil and water bath become
steady for a period of 2-3 minutes record the temperatures separately.
(Quite likely that there may be a difference of 1-20C between the
temperature of oil and the temperature of water in the water bath.
9) Lift up the value rod and suspend it from the thermometer bracket and
start the stop watch simultaneously.
10) Receiving flask is so located that oil strikes the flared mouth and does
not drop directly into the opening, which would cause foaming.
11) When the level of the oil reaches 50 ml. mark in the neck of the flask,
stop the stop-watch.
12) At the same time close the passage of the orifice by keeping the valve
rod at its original position to prevent any overflow of oil.
13) The time elapsed in seconds is the relative viscosity of oil at room
temperature.
14) Switch on the water bath adjust the knob of the regulator in such a way


Page 37


that the temperature of the water bath is a few degree above
temperature i.e. 850C
15) Again, pour oil into the cup and adjust oil level as described earlier.
16) Stir the contents of the bath and dup regularly.
17) When the temperature of sample has become quite steady at the
desired value i.e. at 850C lift the valve rod and suspend it from
the thermometer bracket and start the stopwatch simultaneously.
18) When the level of the oil touches the 50 ml. mark on the neck of the
flask stop the watch. The time elapsed in seconds in the relative
viscosity of oil at 850C.
19) Now take out some hot water from the water bath and add equal
amount of cold water to bring the temperature of the bath slightly
above that of test temperature i.e. 750C.
20) Pour oil into the cup and adjust oil level as described earlier.
21) Stir the contents of the bath and cup regularly.
22) When the temperature of sample has become quite steady at the
desired temperature i.e. 750C find out the relative viscosity of
test sample at this temperature as described earlier.
23) Repeat the procedure to find out relative viscosities of oil at
650C, 550C, 450C, respectively.
24) Record all the results in the tabular form.
25) Construct graph co-relation (a) Viscosity and temperature, (b) Log of
viscosity and temperature and (c) Kinematic viscosity and temperature
.

(d) Log of Kinematic viscosity and temperature, (e) Find viscosity


Page 38


index of the given oil.
Observation table: Redwood viscometer no.:
Sr. Temperature
0

No.

(T) in C

Time
second

in Relative
Viscosity
( ηr )

Kinematic

Log (η r )

Log ( µ )

Viscosity

(µ )
in
centistroke

1

Room Temp.:

2

85

3

75

4

65

5

55

6

45

7

35

8

Calculation:
Graphs were plotted between
1. Relative viscosity ( η r ) against the temperature of sample (T)
2. Log of relative viscosity ( η r ) against the temperature of sample (T)


Page 39


3. Kinematic viscosity ( µ ) against the temperature of sample (T)
4. Log of kinematic viscosity ( µ ) against the temperature of sample (T)
And determine the slope of the graph 1 from the plot and calculate it using the least square method.

Result:- The rate of change of relative viscosity with respect to temperature was found to be equal to


Page 40


EXPERIMENT NO. 7

Aim: Determination of flash point of a combustible liquid by Able’s closed cup
flash point apparatus.

DEFINATION AND OBJECTS:

Theory:The Flash Point of oil may be defined as the minimum temperature to which it must be
heated to give off sufficient vapour to ignite momentarily when a flame of standard dimensions is
brought near the surface of the sample for a prescribed rate in an apparatus of specified dimensions.
This is detected by the appearance of momentary flash upon the application of small flame over the
surface of oil. The Flash Point is defined as closed cup or open cup flash point accordingly as the
apparatus for the determination of flash point of sample is provided with a cover to cover the sample
cup or not.

The mechanism of the appearance of the flash can be explained in the following manner.
Every flammable liquid has a vapour pressure, which is a function of the liquid’s temperature. As the
temperature increase, the vapour pressure increases, as the vapour pressure increases, the concentration
of evaporated flammable liquid in the air increases. Hence, temperature determines the concentration
of its vapour in the air to sustain combustion. The flash point of a flammable liquid is the lowest
temperature at which there can be enough flammable vapour to ignite, when an ignition source is
applied. Oil containing minute quantities of volatile organic substances is liable to flash below the true
flash point of the oil. Although a small flash may be observed in such cases, it should not be confused
with the true flash point, since its intensity does not increase with increase in temperature, as occurred
when the true flash point is reached.


Page 41


Importance of flash point from view of lubricants:

Good lubricating oil should not volatilize under the working temperatures. Even if some
volatilization takes place, the vapour formed should not form inflammable mixture with air under the
condition of lubrication. From this point of view, the flash point of lubricating oil is of vital
importance.

Lubricating oil selected for the job should have a flash point which is reasonably above its
working temperature. This insures the safety against the fire hazards during the storage, transport and
use of the lubricating oil. This test is immense importance for illuminating and lubricating Oils. This
helps in detecting the highly volatile constituents of the oils. If they are highly volatile at ordinary
temperature, the issuing vapors may cause fire hazards. So to ensure safety, certain minimum
temperatures are laid down for fuels and Lubricating Oils below which they should not give off
adequate vapors to make them burn.

OUTLINE OF THE METHOD:

The sample is placed in the oil cup of apparatus and heated at a slow uniform rate about 2
0

C. A small test is directed into the cup at regular intervals, and the lowest temperature at which

application of the test flame causes the vapor to ignite momentary, with a distinct flash inside the cup,
is recorded as Abel’s Flash Point.

Description of the apparatus:

The Abel’s apparatus consists of the following essential parts: -

a.

SAMPLE CUP: This is a cylindrical vessel with a lid. Within the cup, near the top;
there is a sample level mark.

b.

COVER:


Page 42


(A) COVER: The cup is provided with a close fitting cover. The cover is provided with a
thermometer socket, to support an oil test lamp, a movable metal bead. The top of the
cover has three rectangular holes which are covered or uncovered by moving the slide
which has two perforations to allow air to enter the oil up and bring contact between
vapors of the oils and flame of the test-lamp when in open position.

A metal bead, the dimension (4mm) of which represents the size of test flame, is mounted
in the cover. The apparatus is also provided with a stirrer.

HEATING VESSEL: -

The heating vessel consists of copper vessels and placed coaxially, one inside the other.
The space between the two vessels is used as a water jacket. When the oil cup is placed
into the hole at the top of the vessel, it fits into it and leaves an air gap between itself and
outer copper water vessel.

The water can be heated electrically or with a burner or spirit lamp. Thermometer is
provided with the apparatus for the measuring the temperature of the oil sample.

PROCEDURE:

Fill the given sample in such a way that the sample level is exactly up to the mark in the cup.
Fix the cup in to the apparatus and cover with lid. Insert thermometer in the thermometer holder given
in the cup in such a manner that it will not directly touch the lower bottom of the cup and the paddle
stirrer inside the cup. Fill the water bath with the cold water. Close the sliding shutter and light the
standard flame. Adjust the size of flame (4mm diameter) with respect to the metal bead. Stir the oil
using paddle stirrer. Introduce the flame by opening the shutter and check the appearance of the flash.
Now heat the apparatus and set the rate of temperature increase at the rate of 1 to 2 0C per minute.
Check the flash point of given sample at the interval of 3 0C rise in the temperature. Discontinue the
stirring the sample during the introduction of the test flame. On observing a flash, stop the heating

Page 43


process and allow the temperature to decrease. Check the occurrence of a flash at every 1 0C drop in
temperature at which the flash is observed as the flash point of the sample.

Diagram:


Page 44


OBSERVATION: Sample No.:Table 1
S. No.

Increasing
Temperature (0C)

Inference
(No flash or Flash observed)

1
2
3
4
5
6
Table 2
S. No.

Decreasing
Temperature (0C)

Inference

1
2
3
4
5
6

Result: The Flash Point of given Sample oil No.

by Able’ Flash Point apparatus is found to be --

----0C.


Page 45


EXPERIMENT NO. 8
Aim: - Determination of flash point of a combustible liquid by Pensky-Martan’s
closed cup flash point apparatus.

DESCRIPTION OF THE APPARATUS

A Pensky Martin apparatus consists of the following parts:

SAMPLE CUP: Sample cup is a cylindrical vessel, made of brass with a filling mark grooved inside
near the top. It is provided with a lid.
LID: The lid is equipped with the following parts:

i) Stirrer: The stirring device consists of a vertical steel shaft mounted in the center of the cup and
carrying two-bladed brass propellers.

ii) Cover: It has four opening, one for thermometer and the rest for the oxygen entry and exposure of
vapors to test flame.

iii) Shutter: The lid is equipped with a brass shutter operating on the plane of the upper surface of
the cover. The shutter is so shaped and mounted on the lid that when in one position, the holes are
completely closed and when in the other, these orifices are completely opened.

iv) The flame exposure device: The lid is equipped with a pilot lamp with such a mechanism that
its flame operates simultaneously with the shutter. When the shutter is in the ‘open’ position, the
tip is lowered down in the center of the central orifice.

v) Heater: The cup is heated by means of burner or it is electrically heated. The air bath has
cylindrical interior about 4 cm. deep and can be heated by a direct flame or an electric resistance


Page 46


element. The top-plate is also made of metal and mounted with an air gap between it and the air
bath.

Digram:

PROCEDURE:
The cup and its accessories are well cleaned and dried before the test is started. Now
the cup is filled with the oil to be tested up to the level indicated by the filling mark and covered with
the lid. The stirring device, the thermometer and flame exposure device is fixed on the top of the cover.
The cup is now set in the apparatus properly and the thermometer inserted. The test flame is lighted
and adjusted until it is the size of a bead (4mm in diameter). The apparatus is heated so that the heating
rate is maintained, with the help of a rheostat, at 5-6 0C per minute and stirring rate at 1 to 2 rps.


Page 47


Once the heating started, the test flames is applied after each 2 0C rise of temperature
nearer to the sample in the cup by opening the shutter and check the appearance of the flash.
On observing a flash, stop the heating process and allow the temperature to decrease.
Check the occurrence of a flash at every 1 0C drop in temperature. Record the lowest temperature at
which the flash is observed as the flash point of the sample.

OBSERVATION:
Table 1

S. No.

Increasing
Temperature (0C)

Inference

1
2
3
4
5
6

Table 2
S. No.

Decreasing
Temperature (0C)

Inference

1
2
3
4
5
6
RESULT: The Flash Point of given sample determined by Pensky Marten’s Flash Point apparatus
is found to be -------------0C.


Page 48


EXPERIMENT NO. 9
Aim: Determination of flash point of a combustible liquid by Cleveland’s open cup
flash point apparatus.
DEFINITIONS:
I) Flash Point: Defined earlier.
II) Fire point: It is defined as the minimum temperature at which the oil gives off
sufficient vapor which causes it to burn for a period of 5 seconds or more when a test flame is passed
over surface of the oil.

DESCRIPTION OF APPARATUS:
The apparatus consists of a test cup made of without any lid and is equipped with a
handle. The cup is supported by a metal plate known as heating plate. The cup may be heated by an
electric heater mounted below the cup in the apparatus itself
The metal plate has an extension for mounting the test flame and the thermometer
support. The test flame is mounted in such a manner as to permit automatic duplication of the sweep of
the test flame over the sample cup. The size of the flame can be adjusted with respect to the dimension
of metal bead (4 mm).


Page 49


Digram:

PROCEDURE:
The apparatus is thoroughly cleaned and the thermometer is suspended in such a way so
that the bottom of the thermometer bulb just above the bottom of sample cup.
The cup is now filled with sample up to the filling mark grooved on the inner side of
the cup taking care that the surface of the sample is free from bubbles and there is no oil above the
filling mark. The test flame is adjusted to have a flame diameter of about 4 mm. Now move the test
flame over the sample cup and check the appearance of flash over the sample inside the cup.
If no flash observed, increase the temperature of the sample taken and take a flame over the cup after
every 2 0C increases. On observing a flash, stop the heating process and allow the temperature to
decrease. Check the occurrence of a flash at every 1 0C drop in temperature. Record the lowest
temperature at which the flash is observed as the flash point of the sample.


Page 50


OBSERVATION:

Table 1
S. No.

Increasing
Temperature (0C)

Inference

1
2
3
4
5
6

Table 2
S. No.

Decreasing
Temperature (0C)

Inference

1
2
3
4
5
6

Result:
The Flash Point of given sample determined by Cleveland’s apparatus is found to be ----0C.


Page 51


EXPERIMENT NO. 10
Aim : To determine Chemical Oxygen Demand (COD) of a given water sample.
Apparatus:
Reflux apparatus, consisting of a 250 ml.-flat bottomed Borosil flask with ground glass joint and a
condenser.

Regents required:
1. Standard 0.25 N K2Cr2O7 solution
2. Sulfuric acid-silver sulphate reagent:
3. Standard 0.1 N Fe (NH 4) 2SO4. 7H2O solution:
4. Ferroin indicator
5. HgSO4

+2

N

Fe

. SO -4-2

N

Figure: Structure of ferroin indicator

Theory:

In Environmental Chemistry, the Chemical Oxygen Demand (COD) test is commonly

used to measure the amount of organic compounds that is susceptible to oxidation by a strong oxidant
present in water. Any organic matter such as hydrocarons when brought in contact with oxygen is
oxidised to get CO2 and H2O. If the organic matter contains hydrogen, nitrogen, sulphur, etc. in

Page 52


addition to carbon, then the chemical oxidation leads to formation of CO2, H2O, NO2, SO2 and etc. To
get this oxidation, oxygen is required and demanded by this organic material. The possibility of
meeting such demand is through dissolved oxygen in the water. Thus dissolved oxygen in water gets
depleted and survival of the bio-organisms in such water is difficult and sometimes impossible. It is
therefore, essential to check COD of effluent water before letting it out from industry. It is expressed in
milligrams per liter (mg/L), which indicates the mass of oxygen consumed per liter of solution. Older
references may express the units as parts per million (ppm).
The basis for the COD test is that nearly all organic compounds can be fully oxidized to carbon
dioxide with a strong oxidizing agent under acidic conditions. The amount of oxygen required to
oxidize an organic compound to carbon dioxide, ammonia, and water is given by:

This expression does not include the oxygen demand caused by the oxidation of ammonia into
nitrate. The process of ammonia being converted into nitrate is referred to as nitrification. The
following is the correct equation for the oxidation of ammonia into nitrate.

The second equation should be applied after the first one to include oxidation due to
nitrification if the oxygen demand from nitrification must be known. Dichromate does not oxidize
ammonia into nitrate, so this nitrification can be safely ignored in the standard chemical oxygen
demand test.

Using potassium dichromate
Potassium dichromate is a strong oxidizing agent under acidic conditions. (Acidity is usually
achieved by the addition of sulfuric acid.) The reaction of potassium dichromate with organic
compounds is given by:


Page 53


where d = 2n/3 + a/6 - b/3 - c/2. Most commonly, a 0.25 N solution of potassium dichromate is used
for COD determination, although for samples with COD below 50 mg/L, a lower concentration of
potassium dichromate is preferred.
In the process of oxidizing the organic substances found in the water sample, potassium
dichromate is reduced (since in all redox reactions, one reagent is oxidized and the other is reduced),
forming Cr3+. The amount of Cr3+ is determined after oxidization is complete, and is used as an indirect
measure of the organic contents of the water sample.

Importance of Blank Titration
Because COD measures the oxygen demand of organic compounds in a sample of water, it is
important that no outside organic material be accidentally added to the sample to be measured. To
control for this, a so-called blank sample is required in the determination of COD (and BOD, for that
matter). A blank sample is created by adding all reagents (e.g. acid and oxidizing agent) to a volume of
distilled water. COD is measured for both the water and blank samples, and the two are compared. The
oxygen demand in the blank sample is subtracted from the COD for the original sample to ensure a
true measurement of organic matter.

Interference due to Chlorides in waste water
Some samples of water contain high levels of oxidizable inorganic materials which may
interfere with the determination of COD. Because of its high concentration in most wastewater,
chloride is often the most serious source of interference. Its reaction with potassium dichromate
follows the equation:


Page 54


Prior to the addition of other reagents, mercuric sulfate can be added to the sample to eliminate
chloride interference.

Procedure:
Part –I

Determination of COD value of the given waste water sample

Pipette out 10 ml of the wastewater sample in a reflux flask, dilute the sample by 20 ml
with distilled water. Mix well and add 10 ml.of 0.25 N K2Cr2O7 solutions. Drop some pumice stones
and slowly add 30 ml. of H2SO4—AgSO4 reagent while continuously swirling the flask. If the colour
changes to green, add more K2Cr2O7 and or alternatively discard the solution and take a fresh sample
with lesser aliquot. Mix the contents of the flask thoroughly. Connect the flask to the condenser and
slowly heat flask. Reflux for at least 2 hours. Cool and wash down the condenser with distilled water
such that the washings fall in to the flask. Add few drops of Ferroin indicator and titrate the content of
the flask (unreacted K2Cr2O7) with the 0.1 N FAS solution. Near the end point the yellow colour of the
solution gradually fades up to bluish green, continue the titration till bluish green colour changes to
wine red color.

Part-II

Blank Titration

Repeat the above procedure using 10 ml of the distilled water instead of the
Waste water sample for about half hour.

Part-III

Standardization of the FAS solution

Pipette out 10 ml of the K2Cr2O7 solution in a conical flask add few drops of the Ferroin
indicator and titrate with the standard FAS solution till solution changes color from yellow to wine red
colour.


Page 55


Calculation:

Part A] Standardization of FAS solution:

Std. K2Cr2O7 =
N1V1

FAS solution

=

N2V2

N2 =

N1V1
V2

Part B] Determination of COD of given water sample:

The COD value of the sample is calculated as follows:

COD in mg/l =

(V4 -V3 ) × N 2 × 8
×1000
V

Where,
V4= Volume of the FAS run down in the blank Experiment.
V3 = Volume of the FAS run down in the test Experiment.
N2 = Normality of FAS solution
V = Volume of the test sample taken.

Result:

The Chemical oxygen demand of the given waste water sample was found to be equal to

__________ppm.


Page 56


EXPERIMENT NO. 11
Aim: - To estimate saponification value of a given oil sample.
Introduction:
The saponification value of oil is the number of milligrams of potassium hydroxide required to
saponify one gram of the oil or fat under the specified conditions.
The vegetables or animal oils are esters of fatty acids and glycerol. They react with KOH to
form the potassium salts of fatty acids. The saponification value of oil is determined by refluxing a
known quantity of the sample with a known excess of standard KOH solution and determining the
alkali consumed by titrating the unreacted alkali.

It is a measure of the average molecular weight (or

chain length) of all the fatty acids present. As most of the mass of a fat/triester is in the 3 fatty acids, it
allows for comparison of the average fatty acid chain length. The long chain fatty acids found in fats
have low Saponification value because they have a relatively fewer number of carboxylic functional
groups per unit mass of the fat as compared to short chain fatty acids.

Reactions:
CH2OOCC17H35 CH2OH
||
CHOOCC17H35 + 3KOH ------CHOH + 3C17H35COOK
||
CH2OOC17H35
CH2OH

Fat

Stearic acid glycerol potassium steareate


Page 57


The Acid Value of oil is defined as the number of milligrams of potassium hydroxide required
to neutralize the free acid present in 1 gram of the oil.
The presence of mineral acids in oil is so rare that it is almost unnecessary to look for it, unless
the oil is refined in a faulty manner. But, Free organic acids or acidic bodies are always found in Oil,
whether they be pure mineral oils or compounded oils with fatty oils.
Mineral oils (petroleum oil) are contaminated by mineral acids due to improper refining. In
unused refined petroleum oils, the quantity is invariably negligible. Since animal & vegetable oils are
the esters of higher fatty acids which on hydrolysis forms organic acids. Hence animal and vegetable
oils are generally contaminated by organic acids.
Acid present in lubricating oil should be determined because this acid may cause corrosion of
equipment. In good lubricating oil the acid value should be less than 0.1 (<0.1). Increase in acid Value
taken as an indicator of oxidation of the oil and may lead to formation of gum and sludge which spoils
the lubrication process. The Acid Value of fatty oils may vary from 0.2 to 50 and it shows the extent of
hydrolysis of glycerol ester of the oil.

Reaction:
C17H35COOH+KOH ---------------- > C17H35COOK+H2O

Regents required:
(1) Oxalic acid solution (2) KOH solution (3) Oil sample (4) Neutral ethyl alcohol
(5) Phenolphthalein indicator.


Page 58


Procedure:
Part-I: Standardization of KOH Solution

Take 10 ml of oxalic acid solution in conical flask. Add few drops of phenolphthalein indicator.
Titrate this solution with KOH solution from burette till colour changes from colourless to light pink.
Note the end point as V2.
Part-II: - Determination of Acid value of Solution
Weight out accurately about 5 g. of the oil under test into a 250 ml conical flask and add 50 ml
of neutral alcohol. Heat the flask over a water bath for about 30 minutes. Cool the flask and the
contents to room temperature and add a few drops of phenolphthalein indicator. Titrate with the KOH
solution until a faint permanent pink color appears at the end point as V2’.
Part-III: - Blank titration
Repeat the same procedure without taking lubricating oil.

Calculations:
Part-I: Standardization of KOH Solution

Std Oxalic acid

N1V1

Vs

KOH

=

N2V2

N1V1
Normality of KOH i e; N2 = ---------------V2


Page 59


Part-II: - Determination of Acid value of Solution

Normality of KOH x V2’ x eq. wt. of KOH
Acid value =

--------------------------------------------------------Weight of the oil taken in g

Result: The Acid Value of the given oil is found to be_________.


Page 60

Applied chemistry practical manual session 12 13

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